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Nastasia [14]
3 years ago
15

If a calibration curve of absorbance vs procaine hydrochloride concentration (mol L') is linear with the equation y = 240000x fo

r the derivatised procaine, what is the value of the molar absorptivity coefficient (Emolar)? Assume that a cuvette with a pathlength of 3 cm was used. (1 mark) Y=240000 Emolar = _ L molo
Physics
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

8000 L/ mol cm is the value of the molar absorptivity coefficient.

Explanation:

The slope of the absorbance- concentration graph is equal to the product of path length and  molar absorptivity coefficient.

\frac{dA}{dc}=\epsilon l

\frac{dA}{dc}= Rate of change of absorbance with respect to concentration

\epsilon = molar absorptivity coefficient

l = Path-length

Given the slope of graph of absorbance vs procaine hydrochloride concentration.

y = 240000 x (linear equation)

Path length of cuvette = 3 cm

A=240000 c

A=\epsilon c\times l=(\epsilon l)\times c

240000=\epsilon \times 3

\epsilon =\frac{24000}{3}=8000

The units of \epsilon = 8000 l/mol cm

8000 L/ mol cm is the value of the molar absorptivity coefficient.

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Which of the following is a common feature among the four inner planets?
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A spring with a constant k=400n/m shoots a 1. 00kg ball up a frictionless incline after being compressed 0. 150m. what is the ma
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The maximum height reached by the ball is 0.46m.

To find the answer, we have to know about the potential energy of a spring mass system.

<h3>How to find the maximum height reached by the ball?</h3>
  • It is given that,

                       k=400N/m\\m=1kg\\x=0.150m\\h=?\\

  • We have to find the maximum height reached by the ball.
  • Thus, we have the expression for potential energy of a spring mass system and that of gravitational field as,

                             U=\frac{1}{2}kx^2 \\U=mgh

  • Combining both, we get,

                    h=\frac{kx^2}{2mg} =\frac{400*(0.15)^2}{2*1*9.8} =0.46m

Thus, we can conclude that, the maximum height reached by the ball is 0.46m.

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brainly.com/question/26962934

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8 0
1 year ago
A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your
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Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

x = x_{0} + V_{0}t + \frac{1}{2}at^2

Where:

x = Final distance

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a = Acceleration

t = time

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x = 1.66m      

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t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

x = x_{0} + V_{0}t + \frac{1}{2}at^2

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It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

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